It is perhaps worth noting that the elements of $\mathbb Q(\sqrt{5})$ which are integral over $\mathbb Z$ include $\frac {1+\sqrt 5} 2$ (integral in this context means numbers which are roots of monic polynomials with integer coefficients). So this is just a check that the question is correctly posed for its context.
–
Mark BennetJul 1 '12 at 18:51

2 Answers
2

Since $a^2-5b^2\equiv a^2\pmod{5}$, the values of $a^2-5b^2$ must be congruent to either $0$, $1$, or $4$ modulo $5$, since those are the only squares modulo $5$. But neither $2$ nor $-2$ satisfy this condition, so it is impossible to have $a^2-5b^2=2$ or $a^2-5b^2=-2$ with $a,b\in\mathbb{Z}$.